问题
I have many (~ 1 million) irregularly spaced points, P, along a 1D line. These mark segments of the line, such that if the points are {0, x_a, x_b, x_c, x_d, ...}, the segments go from 0->x_a, x_a->x_b, x_b->x_c, x_c->x_d, etc. I also have a y-value for each segment, which I wish to interpret as a depth of colour. I need to plot this line as an image, but there might only be (say) 1000 pixels available to represent the entire length of the line. These pixels, of course, correspond to regularly spaced intervals along the line, say at 0..X1, X1..X2, X2..X3, etc, where X1, X2, X3 are regularly spaced. To work out the colour for each pixel, I need to take an average of all the y-values that fall within the regularly-spaced pixel boundaries, weighted by the length of the segment that falls in that interval. There are also likely to be pixels which do not contain any value in P, and which simply take the colour value defined by the segment that passes across the entire pixel.
This seems like something that probably needs to be done a lot in image analysis. So is there a name for this operation, and what's the fastest way in numpy to calculate such a regularly-spaced set of average y-values? It's a bit like interpolation, I guess, only I don't want to take the mean of just the two surrounding points, but a weighted average of all points within a regular interval (plus a bit of overlap).
[Edit - added minimal example]
So say there are 5 segments along a horizontal line, delimited by [0, 1.1, 2.2, 2.3, 2.8, 4] (i.e. the line goes from 0 to 4). Assume each of the segments takes an arbitrary shading values, for example, we could have the 5 shading values [0,0.88,0.55,0.11,0.44] - where 0 is black and 1 is white. Then if I wanted to plot this using 4 pixels, I would need to create 4 values, from 0...1, 1...2, etc, and would expect the calculation to return the following values for each:
0...1 = 0 (this is covered by the first line segment, 0->1.1)
1...2 = 0.1 * 0 + 0.9 * 0.88 (1 ... 1.1 is covered by the first line segment, the rest by the second)
2...3 = 0.2 * 0.88, 0.1 * 0.55 + 0.5 * 0.11 + 0.2 * 0.44 (this is covered by the second to the fifth line segments )
3...4 = 0.44 (this is covered by the last line segment, 2.8->4)
Whereas if I wanted to fit this data into a 2-pixel-long line, the 2 pixels would have the following values:
0...2 = 1.1 / 2 * 0 + 0.9 / 2 * 0.88
2...4 = 0.2 / 2 * 0.88 + 0.1 / 2 * 0.55 + 0.5 / 2 * 0.11 + 1.2 * 0.44
This seems like the "right" way to do downsampling along a 1d line. I'm looking for a fast implementation (ideally something build-in) for when I have (say) a million points along the line, and only 1000 (or so) pixels to fit them in.
回答1:
As you would expect, there is a purely numpy solution to this problem. The trick is to cleverly mix np.searchsorted, which will place your regular grid on the nearest bin of the original, and np.add.reduceat to compute the sums of the bins:
import numpy as np
def distribute(x, y, n):
"""
Down-samples/interpolates the y-values of each segment across a
domain with `n` points. `x` represents segment endpoints, so should
have one more element than `y`.
"""
y = np.asanyarray(y)
x = np.asanyarray(x)
new_x = np.linspace(x[0], x[-1], n + 1)
# Find the insertion indices
locs = np.searchsorted(x, new_x)[1:]
# create a matrix of indices
indices = np.zeros(2 * n, dtype=np.int)
# Fill it in
dloc = locs[:-1] - 1
indices[2::2] = dloc
indices[1::2] = locs
# This is the sum of every original segment a new segment touches
weighted = np.append(y * np.diff(x), 0)
sums = np.add.reduceat(weighted, indices)[::2]
# Now subtract the adjusted portions from the right end of the sums
sums[:-1] -= (x[dloc + 1] - new_x[1:-1]) * y[dloc]
# Now do the same for the left of each interval
sums[1:] -= (new_x[1:-1] - x[dloc]) * y[dloc]
return new_x, sums / np.diff(new_x)
seg = [0, 1.1, 2.2, 2.3, 2.8, 4]
color = [0, 0.88, 0.55, 0.11, 0.44]
seg, color = distribute(seg, color, 4)
print(seg, color)
The result is
[0. 1. 2. 3. 4.] [0. 0.792 0.374 0.44 ]
Which is exactly what you expected in your manual computation.
Benchmarks
I ran the following set of benchmarks to make sure that both EE_'s solution and mine agreed on the answer, and to check out the timings. I modified the other solution slightly to have the same interface as mine:
from scipy.interpolate import interp1d
def EE_(x, y, n):
I = np.zeros_like(x)
I[1:] = np.cumsum(np.diff(x) * y)
f = interp1d(x, I, bounds_error=False, fill_value=(0, I[-1]))
pix_x = np.linspace(x[0], x[-1], n + 1)
pix_y = (f(pix_x[1:]) - f(pix_x[:-1])) / (pix_x[1:] - pix_x[:-1])
return pix_x, pix_y
And here is the testbench (the method MadPhysicist is just renamed from the distribute function above). The input is always 1001 elements for x and 1000 elements for y. The output numbers are 5, 10, 100, 1000, 10000:
np.random.seed(0x1234ABCD)
x = np.cumsum(np.random.gamma(3.0, 0.2, size=1001))
y = np.random.uniform(0.0, 1.0, size=1000)
tests = (
MadPhysicist,
EE_,
)
for n in (5, 10, 100, 1000, 10000):
print(f'N = {n}')
results = {test.__name__: test(x, y, n) for test in tests}
for name, (x_out, y_out) in results.items():
print(f'{name}:\n\tx = {x_out}\n\ty = {y_out}')
allsame = np.array([[np.allclose(x1, x2) and np.allclose(y1, y2)
for x2, y2 in results.values()]
for x1, y1 in results.values()])
print()
print(f'Result Match:\n{allsame}')
from IPython import get_ipython
magic = get_ipython().magic
for test in tests:
print(f'{test.__name__}({n}):\n\t', end='')
magic(f'timeit {test.__name__}(x, y, n)')
I will skip the data and agreement printouts (the results are identical), and show the timings:
N = 5
MadPhysicist: 50.6 µs ± 349 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
EE_: 110 µs ± 568 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
N = 10
MadPhysicist: 50.5 µs ± 732 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
EE_: 111 µs ± 635 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
N = 100
MadPhysicist: 54.5 µs ± 284 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
EE_: 114 µs ± 215 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
N = 1000
MadPhysicist: 107 µs ± 5.73 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
EE_: 148 µs ± 5.11 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
N = 10000
MadPhysicist: 458 µs ± 2.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
EE_: 301 µs ± 4.57 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
You can see that at smaller output sizes, the numpy solution is much faster, probably because the overhead dominates. However, at larger numbers of break points, the scipy solution becomes much faster. You would have to compare at different input sizes to get a real idea of how the timing works out, not just different output sizes.
回答2:
You can still accomplish this with linear interpolation. Although your function is piecewise constant, you want its average over a whole lot of small intervals. The average of some function f(x) over an interval from a to b is just its integral over that range divided by the difference between a and b. The integral of a piecewise constant function will be a piecewise linear function. So, supposing you have your data:
x = [0, 1.1, 2.2, 2.3, 2.8, 4]
y = [0, 0.88, 0.55, 0.11, 0.44]
Make a function which will give its integral at any value of x. Here the array I will contain the value of the integral at each of your given values of x, and the function f is its linear interpolant which will give the exact value at any point:
I = numpy.zeros_like(x)
I[1:] = numpy.cumsum(numpy.diff(x) * y)
f = scipy.interpolate.interp1d(x, I)
Now evaluating the average over each of your pixels is easy:
pix_x = numpy.linspace(0, 4, 5)
pix_y = (f(pix_x[1:]) - f(pix_x[:-1])) / (pix_x[1:] - pix_x[:-1])
We can check what is in these arrays:
>>> pix_x
array([0., 1., 2., 3., 4.])
>>> pix_y
array([0. , 0.792, 0.374, 0.44 ])
The shading values for your pixels are now in pix_y. These should match exactly the values you gave above in your example.
This should be quite fast even for many, many points:
def test(x, y):
I = numpy.zeros_like(x)
I[1:] = numpy.cumsum(numpy.diff(x) * y)
f = scipy.interpolate.interp1d(x, I,
bounds_error=False, fill_value=(0, I[-1]))
pix_x = numpy.linspace(0, 1, 1001)
pix_y = (f(pix_x[1:]) - f(pix_x[:-1])) / (pix_x[1:] - pix_x[:-1])
return pix_y
timeit reports:
225 ms ± 37.6 ms per loop
on my system when x is of size 1000000 (and y of size 999999). Note that bounds_error=False and fill_value=(0, I[-1]) are passed to interp1d. This has the effect of assuming that your shading function is zero outside of the range of x-values. Also, interp1d does not need the input values to be sorted; in the above test I gave both x and y as arrays of uniform random numbers between 0 and 1. However, if you know for sure that they are sorted, you can pass assume_sorted=True and you should get a speed increase:
20.2 ms ± 377 µs per loop
回答3:
Given your key requirement that you do not want linear interpolation, you should take a look at using scipy.signal.resample.
What this will do is convert your signal into a spectrum, and then into a new time series which is regularly spaced along the x-axis.
Also see this question: How to uniformly resample a non uniform signal using scipy.
来源：https://stackoverflow.com/questions/52312513/numpy-fast-regularly-spaced-average-for-large-numbers-of-line-segments-points